Answer:
The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 24 - 1 = 23
98% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 23 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 2.5
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.02 = $40.98.
The upper end of the interval is the sample mean added to M. So it is 42 + 1.02 = $43.02.
The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.
Answer:
50°
Step-by-step explanation:
The internal angles on a quadrilateral sum to 360.
So x+10+2x+x+3x=360
7x+10=360
7x=350
x=50
Answer:
a) 68.2%
b) 31.8%
c) 2.3%
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 530
Standard Deviation, σ = 119
We are given that the distribution of math scores is a bell shaped distribution that is a normal distribution.
Formula:
a) P(test scores is between 411 and 649)
b) P(scores is less than 411 or greater than 649)
c) P(score greater than 768)
P(x > 768)
Calculation the value from standard normal z table, we have,
The answer would 9 cm. All of the given choices are wrong
since the measurement maybe too big or too small. Like for 9 millimeter, it
would be too small for a clothespin. While 9km is too big for a clothespin,
there is no such thing as a 9km clothespin. While 9 m is also too big. The
approximate size of a clothespin is 3 inches, when we convert it to cm, it will
be 7 cm. So 9 cm is a best approximate.
